Question

A monochromatic source of light operating at 15 kW emits \(2.5 \times 10^{22}\) photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to ________.

• A. Microwave
• B. Infrared
• C. Visible
• D. Ultraviolet

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The total power output of the light source is the product of the number of photons emitted per second and the energy of a single photon. By finding the energy of one photon, we can calculate its wavelength and determine its region in the electromagnetic spectrum.
Step 2: Key Formula or Approach:
1. Power $P = n \times E$, where $n$ is photons per second and $E$ is energy per photon.
2. Energy of a photon $E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E}$.
Step 3: Detailed Explanation:
Given values:
Total power $P = 15\text{ kW} = 15000\text{ W} = 1.5 \times 10^4\text{ J/s}$.
Emission rate $n = 2.5 \times 10^{22}\text{ photons/s}$.
First, calculate the energy $E$ of a single photon:
$P = n \times E \implies E = \frac{P}{n} = \frac{1.5 \times 10^4}{2.5 \times 10^{22}} = 0.6 \times 10^{-18}\text{ J} = 6 \times 10^{-19}\text{ J}$.
Now, calculate the wavelength $\lambda$:
$\lambda = \frac{hc}{E} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6 \times 10^{-19}}$.
$\lambda = \frac{19.8 \times 10^{-26}}{6 \times 10^{-19}} = 3.3 \times 10^{-7}\text{ m}$.
Convert meters to nanometers ($1\text{ m} = 10^9\text{ nm}$):
$\lambda = 3.3 \times 10^{-7} \times 10^9\text{ nm} = 330\text{ nm}$.
The visible light spectrum typically spans from $400\text{ nm}$ (violet) to $700\text{ nm}$ (red). Wavelengths shorter than $400\text{ nm}$ fall into the Ultraviolet (UV) region.
Since $330\text{ nm}<400\text{ nm}$, the radiation belongs to the Ultraviolet region.
Step 4: Final Answer:
The electromagnetic radiation belongs to the Ultraviolet region.